math/check+ help! conic in standard form

Hi guys!

use the graph 4x^2-y^2-8x-4y+16=0 to do the following:

a. write the equation in standard form

I did but could you double check my work?

4x^2-y^2-8x-4y+16=0

4x^2-y^2-8x-4y=-16

4(x^2-2x) -(Y^2+4y) =-16

4[(X^2-2x+1)-1] - [(Y^2+4y+4)-4] =-16

4[(x-1)^2-1] - [(Y+2)^2-4] =-16

4(x-1)^2-4 - (y+2)^2+4 =-16

4(x-1)^2- (y+2)^2= -16

__4(x-1)^2- (y+2)^2= -16__

-16

__(x-1)^2__ __- (y+2)^2 = 1__

-4 ....... 16

I think I did that right! but not sure!

there is another part to it:

locate the center and the vertices of this curve

The center is:

(-2,1)

how would I find out the vertices of the curve?

Thanks for your help!